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Expansion of (1/x) * Series_Reversion( x / ( (1+x)^2 * (1+3*x)^2 ) ).
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%I #10 Mar 22 2024 09:00:33

%S 1,8,86,1064,14289,202488,2980380,45122792,698214548,10993069856,

%T 175546104958,2836384141720,46285381498750,761735217877200,

%U 12628402069223160,210704642400488040,3535494883741420908,59621314428576557664,1009942893735988354296

%N Expansion of (1/x) * Series_Reversion( x / ( (1+x)^2 * (1+3*x)^2 ) ).

%H <a href="/index/Res#revert">Index entries for reversions of series</a>

%F a(n) = (1/(n+1)) * Sum_{k=0..n} 3^k * binomial(2*(n+1),k) * binomial(2*(n+1),n-k).

%o (PARI) my(N=20, x='x+O('x^N)); Vec(serreverse(x/((1+x)^2*(1+3*x)^2))/x)

%o (PARI) a(n) = sum(k=0, n, 3^k*binomial(2*(n+1), k)*binomial(2*(n+1), n-k))/(n+1);

%Y Cf. A002293, A371406.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Mar 21 2024